Thursday, May 13, 2010

Problem 451: Triangle, Median, Congruence, Proportional Segments

Geometry Problem
Click the figure below to see the complete problem 451 about Triangle, Median, Congruence, Proportional Segments.

Problem 451: Triangle, Median, Congruence, Proportional Segments.
See also:
Complete Problem 451
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 11:11 AM
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3 comments:

  1. c .t . e. oMay 13, 2010 at 11:50 AM

    draw CKL//EF, K on BD, L on AB
    ▲BFG ~ ▲BLK, ▲BGE ~ ▲BKC
    => KL = kx, KC = ky, (k koeficent = BF/FL)
    from P 450
    kx/ky = a/c

    x/y = a/c
    -----------------------------------------

    ReplyDelete
  2. AnonymousMay 14, 2010 at 12:23 AM

    A nice solution

    ReplyDelete
  3. Sumith PeirisSeptember 13, 2017 at 10:12 AM

    It's not difficult to see that

    S(ABG) = S(CBG)

    So c.x sin < BFE = a.y sin < BEF

    Hence cx = ay

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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